Question

calculate the ph of a solution prepared by adding 115ml of 0.1 M NaOh to 100ml of 0.1M HNO3 solution.

Answer is 11.84

I don't know to solve it.

Answer 1

Number of moles of NaOH , n = molarity x volume in L

                                           = 0.1M x 0.115 L

                                           = 0.0115 moles

Number of moles of HNO3 is , n' = Molarity x volume in L

                                               = 0.1M x 0.1 L

                                               = 0.01 mol

NaOH + HNO₃ NaNO₃ + H2O

According to the balanced equation,

1 mole of NaOH reacts with 1 mole of HNO₃

0.01 mole of NaOH reacts with 0.01 mole of HNO₃

So 0.0115 - 0.01 = 0.0015 moles of NaOH left unreacted.

Since NaOH (base) & HNO₃ (acid) are strong , both react with each other to form neutral solution.

So after completion of reaction 0.0015 moles of NaOH left unreacted which results basic nature of the solution.

So Concentration of NaOH left in the solution is = number of moles of NaOH left/Total volume in L

                                                                     = 0.0015 mol / ( 0.115+0.100) L

                                                                     = 0.007 M

1 mole of NaOH contains 1 mole of OH^-

So [OH^-] = [NaOH] = 0.007 M

pOH = - log[OH^-]

       = - log 0.007

       = 2.16

We know that pH + pOH = 14

                     pH = 14 - pOH

                          = 14 - 2.16

                          = 11.84